User blog:Gleive/Gleive's U Notation - Part 1 of ? Parts
I originally planned to release this on September 1st, but meh. While going through this notation, you might notice that despite being finished in 2018 (bruh), it doesn't really grow as spectacularly fast or introduces some amazing new stuff. This is just a normal googologist's normal work, and I just want to say that I've tried many thing in googology and this is just a milestone I want to keep. Anyways, enough semi-emotional stuff, let's get into the notation. I mentioned in my previous post that this notation was refined and revisited several times. The original version was an extension on down-arrow notation (Maybe because BEAF is based on up-arrows?), the second version used my own functions, and the third version meddled with polygonal numbers. But the current version is an extension on Chained arrows. Yes, I know that this is based on some other notation, but the "my own functions" didn't really live up to what I expected though. And chained arrows have some kind of meaning to me (Dunno why...). Maybe because my favorite googolism is (Conway's) Tetratri. First, I think we should go through my take on chained arrows. a→b = a^b a→b→c = a^^...^^b with c ^s a→b→c→d = a→b+1→(a→b+1→c-1→d)→d-1 X→b→c→d = X→b+1→(X→b+1→c-1→d)→d-1 (Where X is an arbitary chain) X→1 = X 1→X = 1 a→1→X = a X→1→X'→d = X→X'→d+1 X→1→1→...→1→X' (Where n is the number of 1s) = X→(n→n→...→n)→X' (n number of ns) Now, let's write → as <1>. a→b→c→d = a<1>b<1>c<1>d Two arrows are <2> a<2>b = a<1>a<1>...<1>a with b a's. Note that we can use multiple arrows in one 3<2>3<1>3 = 3<2>(3<2>2<1>2) When we use multiple arrows in one (Like 3<4>4<5>2), the degeneration rule needs some modification : For four or more entires, follow the original. X→b→c→d = X→b+1→(X→b+1→c-1→d)→d-1. Where the arrow mark denotates an arbitary number of arrows. For three entires, follow a→b→c = a→(a→b-1→c)→c-1. The arrow mark denotates an arbitary number of arrows. The multiple arrows decay only when two numbers and the said arrow is in a single chain. For example, in 3<3>3<1>2, 3<3>3 doesn't turn into 3<2>3<2>3, but in (3<3>3)<1>2, it turns into 3<2>3<2>3. So that's my take on chained arrows so far. Define a as the first entry, t as the third last entry, p as the second last entry and q as the last entry. For a chain where there are only single arrows : Rule A1 : One entry - v© = a Rule A2 : Two entries - v© = a^b Rule A3 : Three entires - v© = a^^...^^b (c times) Rule A4 : Arbitary amount - X→t+1→(X→t+1→p-1→q)→q-1 For a chain where there are multiple arrows : If there is only ac in a chain : ac = aaa...aa c times. With mixed arrows, three entries - abc = a(ab-1c)c-1 With mixed arrows, arbitary amounts - Xt+1(Xt+1p-1q)q-1 Ones at the end are cropped off. If a = 1, then v© = 1. If b = 1, them v© = a. If there are a string of 1's of lenght n, group them in a single chain and convert all the 1s to n's. eg) 3<1>1<1>1<1>1<1>1<1>3 = 3<1>(4<1>4<1>4<1>4)<1>3 Examples of valid forms on current notation : 3<4>2 4<3>3<5>2 3<1>3<1>10 Since the notation name is gleive's "U" notation, why not add a U at the beginning of each chain? U3<4>2 U4<3>3<5>2 U3<1>3<1>10 In action : Conway's tetratet = U4<3>2 Tritri = U3<2>2 This is just the bare bones of the notation, about 1.5% the way there. Part 2 will come soon. Category:Blog posts